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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[algebraic varieties|HV6YBS0YDH]] over an algebraically closed field $k$.\nConsider pairs $(U, \\varphi_U)$ where $U \\subseteq X$ is a nonempty [[open subset|02BH3MCBQB]] and $\\varphi_U \\colon U \\to Y$ is a [[morphism|CDS1VWB7A3]].\nTwo pairs $(U, \\varphi_U)$ and $(V, \\varphi_V)$ are equivalent if $\\varphi_U = \\varphi_V$ on $U \\cap V$.\nA *rational map* $\\varphi \\colon X \\dashrightarrow Y$ is an equivalence class of such pairs.\nIt is *dominant* if for some, hence every, pair $(U, \\varphi_U)$ in the class, the image $\\varphi_U(U)$ is [[dense|0BSJSX7ME2]] in $Y$.\n\n## Remarks\n\nThis relation is an [[equivalence relation|0G8VEZ3WTK]].\nThe key fact is that two morphisms $U \\to Y$ that agree on a nonempty open subset of $U$ are equal.\nThe reason is that the set where they agree is closed, while a nonempty open subset of the irreducible space $U$ is dense.\nThe union of the domains of all pairs in the class is the largest open set on which $\\varphi$ is defined, its domain of definition.\n\nExample: $\\varphi \\colon \\mathbb{P}^2 \\dashrightarrow \\mathbb{P}^1$, $[x : y : z] \\mapsto [x : y]$, is a rational map.\nIt is a morphism on $U = \\mathbb{P}^2 \\setminus \\set{[0 : 0 : 1]}$, and it cannot be extended to a morphism on all of $\\mathbb{P}^2$.\nIt is dominant, since it is surjective onto $\\mathbb{P}^1$.\n\nNon-example: on $X = k$, the assignment $x \\mapsto 1/x$ for $x \\ne 0$ and $0 \\mapsto 0$ is a function $X \\to k$ but not a morphism, so it is not a pair in the definition.\nIts restriction to $k \\setminus \\set{0}$ is a morphism and defines a rational map $k \\dashrightarrow k$, whose domain of definition is $k \\setminus \\set{0}$.\n\nRational maps can only be composed when the first one is dominant; otherwise the image may lie outside the domain of the second.\nA rational map $X \\dashrightarrow \\mathbb{A}^1$ is the same as a rational function on $X$.\n",
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